Irodov 3.3: A Leak Rate Read from Approach – Amit & Deepali Sinha Foundation

Problem Irodov 3.3

Two small spheres of mass $m$ each carry equal charges and hang from a common point on insulating threads of length $l$ . They are a distance $x$ apart, with $x$ much smaller than $l$ . As charge slowly leaks away from both spheres, they move toward each other with relative speed $v = a/\sqrt{x}$ , where $a$ is a constant. Find the rate $dq/dt$ at which each sphere loses charge.

The two spheres of mass m on threads of length l from one point, a distance x apart, each carrying the same charge (drawn positive). The angle is drawn larger than the problem's x l. — The two spheres of mass $m$ on threads of length $l$ from one point, a distance $x$ apart, each carrying the same charge (drawn positive). The angle is drawn larger than the problem’s $x \ll l$ .

Given

$m$
$l$
separation $x \ll l$
approach speed $v = a/\sqrt{x}$

Find

the leak rate

dq/dt

from each sphere

Solution

Step 1The charge that holds the spheres at separation x

The spheres approach slowly, so at each moment each one is in equilibrium under the tension $\vect{T}$ , its weight $m\vect{g}$ and the electric repulsion $\vect{F}$ of the other. Let $\theta$ be the angle of a thread from the vertical and $q$ the charge on each sphere.

The right sphere: tension T along the thread at angle θ to the vertical, weight mg, and the horizontal repulsion F of the other sphere, lengths in proportion for the drawn angle. — The right sphere: tension $\vect{T}$ along the thread at angle $\theta$ to the vertical, weight $m\vect{g}$ , and the horizontal repulsion $\vect{F}$ of the other sphere, lengths in proportion for the drawn angle.

Balancing the horizontal and the vertical components and dividing one by the other:

\begin{aligned} T\sin\theta &= F\\ T\cos\theta &= mg\\ \tan\theta &= \frac{F}{mg} \end{aligned}

For $x \ll l$ the angle is small, and each sphere sits $x/2$ from the vertical through the support:

\tan\theta \approx \sin\theta = \frac{x/2}{l} = \frac{x}{2l}

With Coulomb’s law $F = q^2/(4\pi\varepsilon_0x^2)$ the balance becomes

\begin{aligned} \frac{q^2}{4\pi\varepsilon_0x^2} &= \frac{mgx}{2l}\\ q^2 &= \frac{2\pi\varepsilon_0 mg}{l}\,x^3 \end{aligned}

q = \sqrt{\frac{2\pi\varepsilon_0 mg}{l}}\;x^{3/2}

(1)

Step 2Differentiating along the approach

As the charge leaks the spheres pass through a sequence of such equilibria, so (1) holds at every instant. Differentiating it with respect to time:

\frac{dq}{dt} = \frac{3}{2}\sqrt{\frac{2\pi\varepsilon_0 mg}{l}}\;x^{1/2}\,\frac{dx}{dt} \tag{2}

The spheres approach each other, so the separation shrinks at the given rate:

\frac{dx}{dt} = -v = -\frac{a}{\sqrt{x}}

Substituting this into (2), the factor $\sqrt{x}$ cancels:

\frac{dq}{dt} = -\frac{3}{2}\sqrt{\frac{2\pi\varepsilon_0 mg}{l}}\;x^{1/2}\,\frac{a}{x^{1/2}}

\left\lvert\frac{dq}{dt}\right\rvert = \frac{3}{2}\,a\sqrt{\frac{2\pi\varepsilon_0 mg}{l}}

(3)

Because $dx/dt < 0$ , (2) makes $dq/dt$ negative: the charge on each sphere decreases, and at this constant rate; the given law of approach is exactly the one a steady leak produces.

Check

Units: $\varepsilon_0 mg/l$ is in $(\text{C}^2/\text{N\,m}^2)(\text{N})/\text{m} = \text{C}^2/\text{m}^3$ , so its root is in C/m $^{3/2}$ , and $a$ , from $v = a/\sqrt{x}$ , is in m $^{3/2}$ /s; the product is in C/s, a rate of charge. Sense: a heavier sphere or a shorter thread needs more charge for the same $x$ by (1), so the same approach must shed charge faster, as the answer says.

Remark

The quasi-static picture fails as the spheres close in. Their relative acceleration is

\frac{dv}{dt} = \frac{dv}{dx}\,\frac{dx}{dt} = \left(-\frac{a}{2x^{3/2}}\right)\left(-\frac{a}{x^{1/2}}\right) = \frac{a^2}{2x^2}

which grows without bound as $x$ shrinks; once it is no longer small against the horizontal force per unit mass in the balance, $g\tan\theta \approx gx/(2l)$ , equilibrium at each instant is a poor description. The constant leak, through $q \propto x^{3/2}$ , is what makes the approach speed rise as $1/\sqrt{x}$ .